Problem: You have found the following ages (in years) of all 4 tigers at your local zoo: $ 14,\enspace 5,\enspace 8,\enspace 18$ What is the average age of the tigers at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 4 tigers at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{14 + 5 + 8 + 18}{{4}} = {11.3\text{ years old}} $ Find the squared deviations from the mean for each tiger. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $14$ years $2.7$ years $7.29$ years $^2$ $5$ years $-6.3$ years $39.69$ years $^2$ $8$ years $-3.3$ years $10.89$ years $^2$ $18$ years $6.7$ years $44.89$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{7.29} + {39.69} + {10.89} + {44.89}} {{4}} $ $ {\sigma^2} = \dfrac{{102.76}}{{4}} = {25.69\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{25.69\text{ years}^2}} = {5.1\text{ years}} $ The average tiger at the zoo is 11.3 years old. There is a standard deviation of 5.1 years.